Chapter 21 Electromagnetic Induction and Faraday’s Law Induced EMF Almost 200 years ago, Faraday looked for evidence that a magnetic field would induce an electric current with this apparatus: He found no evidence when the current through the lefthand loop was steady, but did see a current induced in the right-hand loop when the switch was turned on or off. In addition, a current will be induced in a wire loop if a magnet is moved through the loop, but not when the magnet is held steady. Faraday’s Law of Induction; Lenz’s Law The induced emf in a wire loop is proportional to the rate of change of magnetic flux through the loop. Magnetic flux: Unit of magnetic flux: weber, Wb. 1 Wb = 1 T·m2 Faraday’s law of induction: The emf induced in a circuit is equal to the rate of change of magnetic flux through the circuit: For coil with N number of turns: The negative sign gives the direction of the induced emf. Lenz’s Law: An induced emf is always in a direction that opposed the original change in flux that caused it. Magnetic flux will change if the area of the loop changes, and also with the angle between the loop and the field. EMF Induced in a Moving Conductor A uniform movable rod is resting on a U shaped conductor. If the rod is made to move at speed 𝑣, it travels a distance of ∆𝑥 = 𝑣∆𝑡 in time ∆𝑡. Hence by Faraday’s law: ∆Φ𝐵 𝐵Δ𝐴 𝐵𝑙𝑣Δ𝑡 𝜀= = = Δ𝑡 Δ𝑡 Δ𝑡 = 𝐵𝑙𝑣 You can see using right hand rule that electrons on the rod experiences a 𝐵𝑞𝑣 force such that they move to the top of the rod. Thus the induced current is clockwise (to counter the increasing flux.) Changing Magnetic Flux Produces an Electric Field A changing magnetic flux induces an electric field; this is a generalization of Faraday’s law. The electric field will exist regardless of whether there are any conductors around. 𝐸 =𝑣×𝐵 Where 𝑣 is the speed of the charged particle or object. Electric Generators A generator is the opposite of a motor—it transforms mechanical energy into electrical energy. This is an ac generator: The axle is rotated by an external force such as falling water or steam. Rotating axle causes change in area of the coil, thus changing flux. This produces electric current in the circuit Transformers and Transmission of Power A transformer consists of two coils, either interwoven or linked by an iron core. A changing emf in one induces an emf in the other. The ratio of the emfs is equal to the ratio of the number of turns in each coil. P=IV The figure aside shows a Step Up transformer. (21-6) Inductance Mutual inductance: a changing current in one coil will induce a current in a second coil. And vice versa; note that the constant M, known as the mutual inductance, is the same: Unit of inductance: the henry, H. 1 H = 1 V·s/A = 1 Ω·s A transformer is an example of mutual inductance. A changing current in a coil will also induce an emf in the same coil (Lenz’s law): Here, L is called the selfinductance Problem 1: An 18.5 cm diameter loop of wire is initially oriented perpendicular to 1.5 T magnetic field. The loop is rotated so that its plane is parallel to the field direction in 0.2s. What is the average induced emf in the loop? Solution: ∆Φ𝐵 𝐴𝐵∆𝑐𝑜𝑠𝜃 𝜀=− =− ∆𝑡 ∆𝑡 𝜋(0.0925)2 × 1.5 × (𝑐𝑜𝑠90 − 𝑐𝑜𝑠0) = = 0.20𝑉 0.2 Problem 2: A circular loop in the plane of a paper lies in a 0.65T magnetic field pointing into the paper. The loop’s diameter changes from 20cm to 6 cm in 0.5 sec. What is (a) direction of induced current? (b) average induced emf? (c) average induced current if the coil resistance is 2.5 Ohms? Solution: (a)Flux is decreasing, so current produce will be to increase the flux, i.e, the field should point into the paper. Hence current should be clockwise. (b) 𝜀 = ∆Φ𝐵 − ∆𝑡 (c) 𝐼 = 𝜀 𝑅 = 𝐵∆𝐴 − ∆𝑡 = 0.0015 𝐴 =0.0037V Problem 3: A transformer has 330 primary turns and 1240 secondary turns. The input voltage is 120V and the output current is 15A. What are the output voltage and input current? Solution: (a) (b) 𝐼𝑠 𝐼𝑃 = 𝑁𝑃 𝑁𝑆 𝑉𝑠 𝑁𝑠 = 𝑉𝑃 𝑁𝑃 1240 𝑉𝑠 = × 120𝑉 = 450.9𝑉 330 1240 𝐼𝑃 = × 15𝐴 = 56𝐴. 330 Problem 4: Determine a formula for the self inductance 𝐿 of a long tightly wrapped solenoid coil of length 𝑙 and cross sectional area A, that contains N turns of wire. Solution: ∆Φ𝐵 Δ𝐼 𝜀 = −𝑁 = −𝐿 Δ𝑡 Δ𝑡 ΔΦ𝐵 𝑆𝑜, 𝐿 = 𝑁 Δ𝐼 We know that for a solenoid, 𝐵 = 𝜇0 𝑁𝐼/𝑙, and Φ𝐵 = 𝐵𝐴. 𝜇0 𝑁𝐼𝐴 Thus, Φ𝐵 = 𝑙 𝜇 𝑁Δ𝐼.𝐴 So, ΔΦ𝐵 = 0 𝑙 ΔΦ𝐵 Hence, 𝐿 = 𝑁 Δ𝐼 = 𝜇0 𝑁2 𝐴 𝑙 LR Circuit When the switch is closed, current starts to flow, but inductor opposes this flow by producing induced emf (Lenz’s law). Hence current will increase less rapidly. The current reaches steady value when there is no more emf in the inductor (I is no longer changing.) If the circuit is then shorted across the battery, the current will gradually decay away. © 2014 Pearson Education, Inc. AC Circuits and Reactance Resistors, capacitors, and inductors have different phase relationships between current and voltage when placed in an ac circuit. The current through a resistor is in phase with the voltage. The current through an inductor lags the voltage by 90°. The current reaches the peak (or trough) ¼ cycle after the voltage does. © 2014 Pearson Education, Inc. In a capacitor, the current leads the voltage by 90°. Reactance of a Circuit Both the inductor and capacitor have an effective resistance (ratio of voltage to current), called the reactance. Inductor: Capacitor: Note that both depend on frequency. (21-11b) (21-12b) LRC Series AC Circuit Analyzing the LRC series AC circuit is complicated, as the voltages are not in phase—this means we cannot simply add them. Furthermore, the reactances depend on the frequency. We calculate the voltage (and current) using what are called phasors—these are vectors representing the individual voltages. Here, at t = 0, the current and voltage are both at a maximum. As time goes on, the phasors will rotate counterclockwise. Some time t later, the phasors have rotated. The voltages across each device are given by the x-component of each, and the current by its x-component. The current is the same throughout the circuit. We find from the ratio of voltage to current that the effective resistance, called the impedance, of the circuit is given by: Phase angle is calculated as: ∅= 𝑡𝑎𝑛−1 𝑋𝐿 − 𝑋𝑐 𝑅 Resonance in AC Circuits The rms current in an ac circuit is: Clearly, Irms depends on the frequency. We see that Irms will be a maximum when XC = XL; the frequency at which this occurs is This is called the resonant frequency. Also, average Power dissipated: 𝑃𝑎𝑣 = 𝑉𝑟𝑚𝑠 𝐼𝑟𝑚𝑠 𝑐𝑜𝑠∅ Problem 5: An inductance coil operates at 240V and 60 Hz. It draws 12.2A. What is the coil’s inductance? Solution: 𝑉 = 𝐼𝑋𝐿 𝑋𝐿 = 2𝜋𝑓𝐿 Hence, 𝑉 240 𝐿= = = 5.22 × 10−2 𝐻 2𝜋𝐹𝐼 2𝜋(60)(12.2) Problem 6: Determine the total impedance, phase angle, and rms current in an LRC circuit connected to a 10kHz, 725V(rms) source if L =28mH, R=8.7 k Ohm, and C=6250pF. Solution: (a) (b) ∅ = −1 𝑋𝐿 −𝑋𝑐 𝑡𝑎𝑛 𝑅 (c) 𝐼𝑟𝑚𝑠 = © 2014 Pearson Education, Inc. 𝑉𝑟𝑚𝑠 𝑍 Problem 7: The variable capacitor in the tuner of an AM radio has a capacitance of 2800pF when the radio is tuned to a station at 580kHz. (a) What must be the capacitance for a station at 1600kHz? (b) What is the inductance? Solution: a). 𝑓 = 1 2𝜋 Hence, 1 𝐿𝐶 𝑓580 𝑓1600 = 𝐶1600 𝐶580 So, 580 2 1600 𝐶1600 = 367.9𝑝𝐹 b) 𝑓 = 1 2𝜋 So, 580 × 1 𝐿𝐶 103 = 1 2𝜋 1 𝐿×2800×10−12 𝐿 = 2.7 × 10−5 𝐻 = 𝐶1600 2800×10−12